i.e. 3X is common in all terms in the 1st equation….

Now to factor the equation, a simple and easy way is

step 1. multiply the co-efficient of (X^2) with the constant term in the equation.

step 2. find two numbers such that, the sum of the numbers gives the co-efficient of the (X) term and the product of them gives the value found in step 1.

step 3: split the (X) term in terms of the two numbers jus found….

step 4: combine all the common terms to get the factor

ur examples worked out….

3X^3 + 12X^2 – 6X = 3X(X^2 + 4X – 2)

2.

X^2 – 5X + 6

= X^2 – 3X – 2X + 6

(because 6 can be written as (-3) * (-2) )

= X( X – 3) – 2 (X – 3)

(TAKE x COMMON FROM 1ST 2 TERMS AND TAKE A TERM COMMON FROM THE LAST TWO SUCH THAT THE RESULT IS THE SAME AS THE FIRST BRACKETED TERM, IN THIS CASE, (X -3), TO GET THAT WE TAKE (-2) COMMON )

= (X -3)(X – 2)

(SINCE (X – 3) IS COMMON IN BOTH TERMS IN THE LAST STEP, WE TAKE THAT AS COMMON NOW)

One eg. of my own….

X^2 + 2X – 15

= X^2 + 5X – 3X – 15

= X(X + 5) – 3(X + 5)

=(X + 5)(X – 3)

For Example, if you multiply 3X and X^2 + 4X – 2, you would get 3X^3 + 12X^2 – 6X

Here’s how to do the 1st one:

3X^3 + 12X^2 – 6X

Try to find the largest numbers that you can divide all three of the numbers by.

3X^3, 12X^2, and 6X can all be divided 3X, right? 3X^3 divided by 3X would be X^2, 12X^2 divided by 3X would be 4X, and 6X divided by 3X would be 2.

Now write the number you divided them by (in this case, 3X)outside a () and write the numbers you got when you divided them (in this case,X^2, 4X, and 2) inside the ().

3X(X^2 + 4X – 2)

Here’s how to do the 2nd one:

X^2 – 5X + 6

This one is a little different from the first one, and solving it is also different, but you’re still looking for two things that multiply to make this.

Because of the way it is written, you need to know that the answer will definitely be in

(X -or + something)(X – or + something) format. You have to figure out if its a – or + and what the “somethings” are.

Try to find two numbers that add up to the coefficient of the variable in the middle of the original expression (in this case, 5) and also multiply to make the last number of the expression (in this case, 6). 2 and 3 add up to make 5 and also multiply to make 6.

Put the numbers in the ().

(X- or + 2)(X-or + 3)

Look back at the original expression to figure out if the signs are – or +. The middle number of the expression is a -5X, and the only way 2 and 3 could make -5 is if one of them was negative and the other one was subtracted from it. -2-3= -5, and -3-2= -5. The last number of the expression is +6, so if both the 2 and 3 are negative, 6 can be positive (two negative numbers multiplied make a positive number).

Put the signs in the ().

(X-2)(X-3)

the Greast Common Factor is 3x. So factor out 3x.

3x ( (3x^3)/3x + (12x^2)/3x – (6x)/3x)

3x (x^2 + 4x – 2)

standar form:

ax^2 + bx + c

a = 1

b = -5

c = 6

x^2 – 5x + 6

to factor, use AC method.

AC method requires you to find two numbers that multiply to AC and add to B.

AC means A * C.

A * C = 1 * 6 = -6

B = -5

so find two numbers that multiply to 6 and add to -5

the two numbers are -2 and -3

since the leadint coefficient is 1, plug the numbers in factor form (x + #) (x + #)

so (x – 2) (x – 3)

hope this helps